Constructing Householder Reflections

Build a Reflection from Its Normal

A reflection across a line in the plane or a plane in three-dimensional space can be specified by the direction perpendicular to it. In \(\mathbb{R}^n\), the corresponding object is a hyperplane through the origin. Its unit normal tells us which component of a vector must change sign: the component parallel to the hyperplane stays fixed, while the component perpendicular to it is reversed. This description leads directly to the Householder matrix.

Work over \(\mathbb{R}\), with vectors written as columns. Let \(\mathbf{u}\in\mathbb{R}^n\) be a unit vector, so \(\mathbf{u}^{\top}\mathbf{u}=1\). The hyperplane with normal \(\mathbf{u}\) is \(W=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{u}^{\top}\mathbf{x}=0\}\). The condition says exactly that \(\mathbf{x}\) is perpendicular to \(\mathbf{u}\). For any \(\mathbf{x}\in\mathbb{R}^n\), its component in the normal direction is \((\mathbf{u}^{\top}\mathbf{x})\mathbf{u}\); subtracting that component leaves a vector in \(W\).

To derive its formula, write an arbitrary vector as the sum of its hyperplane component and its normal component:

The first component is in \(W\), because its dot product with \(\mathbf{u}\) is \(\mathbf{u}^{\top}\mathbf{x}-(\mathbf{u}^{\top}\mathbf{x})(\mathbf{u}^{\top}\mathbf{u})=0\). A reflection keeps the first component and negates the second. Thus its output is \(\mathbf{x}-2(\mathbf{u}^{\top}\mathbf{x})\mathbf{u}\). Since \((\mathbf{u}^{\top}\mathbf{x})\mathbf{u}=(\mathbf{u}\mathbf{u}^{\top})\mathbf{x}\), this gives the matrix formula.

The unit-length condition matters: it makes \((\mathbf{u}^{\top}\mathbf{x})\mathbf{u}\) the normal component itself. The matrix \(\mathbf{u}\mathbf{u}^{\top}\) is \(n\times n\), so \(\mathbf{H}\) acts on column vectors in \(\mathbb{R}^n\).

Why a Householder Matrix Is Symmetric and Orthogonal

Proof. First, \((\mathbf{u}\mathbf{u}^{\top})^{\top}=\mathbf{u}\mathbf{u}^{\top}\), so taking the transpose of the formula gives \(\mathbf{H}^{\top}=\mathbf{I}_n-2\mathbf{u}\mathbf{u}^{\top}=\mathbf{H}\). For the product, use the unit-length assumption to calculate \((\mathbf{u}\mathbf{u}^{\top})(\mathbf{u}\mathbf{u}^{\top})=\mathbf{u}(\mathbf{u}^{\top}\mathbf{u})\mathbf{u}^{\top}=\mathbf{u}\mathbf{u}^{\top}\). Therefore \(\mathbf{H}^2=(\mathbf{I}_n-2\mathbf{u}\mathbf{u}^{\top})^2 =\mathbf{I}_n-4\mathbf{u}\mathbf{u}^{\top}+4\mathbf{u}\mathbf{u}^{\top} =\mathbf{I}_n\). Since \(\mathbf{H}^{\top}=\mathbf{H}\), this also gives \(\mathbf{H}^{\top}\mathbf{H}=\mathbf{I}_n\), which is the defining test for orthogonality. \(\square\)

The equation \(\mathbf{H}^2=\mathbf{I}_n\) expresses that reflecting twice restores every vector. The formula also verifies the geometric description directly. If \(\mathbf{x}\in W\), then \(\mathbf{u}^{\top}\mathbf{x}=0\), so \(\mathbf{H}\mathbf{x}=\mathbf{x}\). For the normal vector itself, \(\mathbf{H}\mathbf{u}=\mathbf{u}-2(\mathbf{u}^{\top}\mathbf{u})\mathbf{u}=-\mathbf{u}\). Thus the hyperplane is fixed and the normal direction is reversed.

Constructing and Checking the Matrix

Common Mistake: Forgetting to Normalize

The formula \(\mathbf{I}_n-2\mathbf{u}\mathbf{u}^{\top}\) requires a unit normal. For the line \(x+y=0\), using the unnormalized vector \((1,1)^{\top}\) as though it were unit gives the incorrect matrix

This is not an orthogonal matrix: its transpose times itself is its square, which equals \(\begin{bmatrix}5&4\\4&5\end{bmatrix}\), not \(\mathbf{I}_2\). The correction is to normalize first. The unit normal is \((1/\sqrt{2},1/\sqrt{2})^{\top}\), so the correct matrix is

Its square is \(\mathbf{I}_2\); it fixes \((1,-1)^{\top}\), which lies in the line, and sends the unit normal to its negative. The failure in the incorrect calculation is not a rounding issue: the unnormalized vector makes the normal component too large.

What the Construction Guarantees

The construction gives a useful set of checks whenever a reflection matrix is requested. The normal vector must be nonzero so it can be normalized. The hyperplane is the set of vectors whose dot product with that normal is zero. The resulting matrix is symmetric and orthogonal, and applying it twice gives the identity. These properties are consequences of the formula and the unit-normal condition, rather than extra assumptions to impose on the answer.

There is also a useful distinction between choosing a normal direction and choosing a particular unit normal. Replacing \(\mathbf{u}\) by \(-\mathbf{u}\) does not change the matrix, because \((-\mathbf{u})(-\mathbf{u})^{\top}=\mathbf{u}\mathbf{u}^{\top}\). Thus either orientation of the same normal line gives the same reflection. The essential data are the hyperplane and its perpendicular direction, not the sign used to represent that direction.